Chapter 2: Polynomials - Practice worksheet with complete answer key
According to your class and board
- Board: CBSE
- Class/Year: 9
- Subject: Mathematics
- Book + Publisher: Mathematics (NCERT)
- Chapter: Polynomials
- Difficulty: Standard
- Goal: Practice
A1. Define a polynomial in one variable.
A2. What is the degree of the polynomial 4x³ + 3x² − 7x + 2?
A3. Give an example of a monomial of degree 5.
A4. What is the zero of the polynomial p(x) = x − 5?
A5. Is √x + 1 a polynomial? Why or why not?
B1. Classify the following as monomials, binomials, or trinomials: (a) 3x² (b) x² + x + 1 (c) 4y − 7
B2. Find the value of p(x) = x² − 3x + 2 at x = −1.
B3. Verify whether x = 2 is a zero of p(x) = x² − 4.
B4. If p(x) = 2x² + 3x − 5, find p(0) and p(1).
B5. Write the coefficient of x² in the polynomial 5 − 3x² + 7x³.
B6. State the Remainder Theorem.
B7. Find the remainder when p(x) = x³ − 2x² + x − 1 is divided by (x − 1).
B8. Using the Factor Theorem, check whether (x + 1) is a factor of x³ + x² + x + 1.
B9. What is a zero polynomial? What is its degree?
B10. Find the zeroes of the polynomial p(x) = x² − 9.
C1. The cost of manufacturing x items is given by C(x) = 2x² − 5x + 100. Find the cost of manufacturing 10 items.
C2. If (x − 2) is a factor of x³ − kx² + 4x − 8, find the value of k.
C3. Factorise: x² − 7x + 12 using the factor theorem.
C4. A rectangular garden has length (x + 3) metres and breadth (x − 2) metres. Express its area as a polynomial and find the area when x = 5.
C5. Find the remainder when 2x³ + 3x² − 5x + 7 is divided by (x + 2) using the Remainder Theorem.
C6. Factorise: 6x² + 5x − 6 by splitting the middle term.
C7. Verify that x = 1, −1, and 3 are zeroes of x³ − 3x² − x + 3.
C8. If p(x) = x³ − 3x² + kx + 5 leaves remainder 11 when divided by (x − 2), find k.
D1. Using suitable identity, factorise: 27x³ + 64y³. Show all steps. (3 marks)
D2. If (x + a) is a factor of both x² + px + q and x² + mx + n, prove that a = (n − q)/(m − p). (3 marks)
D3. Factorise: x³ − 6x² + 11x − 6 using the Factor Theorem. Verify your answer. (3 marks)
D4. Evaluate 103³ using a suitable algebraic identity. (2 marks)
D5. If x + 1/x = 6, find the value of x² + 1/x². (2 marks)
E1. A student says the degree of the polynomial 5 is 5. Is this correct?
E2. Is (x² − 1)/(x + 1) a polynomial? Explain.
A1. A polynomial in one variable is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where aᵢ are constants and n is a non-negative integer.
A2. Degree = 3 (highest power of x).
A3. Example: 2x⁵ (single term, degree 5).
A4. Zero is x = 5 (set x − 5 = 0).
A5. No, because √x = x^(1/2), and the exponent is not a whole number.
B1. (a) Monomial (b) Trinomial (c) Binomial.
B2. p(−1) = (−1)² − 3(−1) + 2 = 1 + 3 + 2 = 6.
B3. p(2) = 4 − 4 = 0. Yes, x = 2 is a zero.
B4. p(0) = 0 + 0 − 5 = −5. p(1) = 2 + 3 − 5 = 0.
B5. Coefficient of x² is −3.
B6. If a polynomial p(x) is divided by (x − a), the remainder is p(a).
B7. Remainder = p(1) = 1 − 2 + 1 − 1 = −1.
B8. p(−1) = −1 + 1 − 1 + 1 = 0. Yes, (x + 1) is a factor.
B9. The zero polynomial is 0. Its degree is undefined (or not defined).
B10. x² − 9 = (x − 3)(x + 3). Zeroes are x = 3 and x = −3.
C1. C(10) = 2(100) − 5(10) + 100 = 200 − 50 + 100 = ₹250.
C2. Since (x − 2) is a factor, p(2) = 0. 8 − 4k + 8 − 8 = 0 → 8 − 4k = 0 → k = 2.
C3. Factors of 12: try x = 3 → 9 − 21 + 12 = 0 ✓. Try x = 4 → 16 − 28 + 12 = 0 ✓. So x² − 7x + 12 = (x − 3)(x − 4).
C4. Area = (x + 3)(x − 2) = x² + x − 6. When x = 5: 25 + 5 − 6 = 24 sq. metres.
C5. Remainder = p(−2) = 2(−8) + 3(4) − 5(−2) + 7 = −16 + 12 + 10 + 7 = 13.
C6. 6x² + 5x − 6 = 6x² + 9x − 4x − 6 = 3x(2x + 3) − 2(2x + 3) = (3x − 2)(2x + 3).
C7. p(1) = 1 − 3 − 1 + 3 = 0 ✓, p(−1) = −1 − 3 + 1 + 3 = 0 ✓, p(3) = 27 − 27 − 3 + 3 = 0 ✓. All verified.
C8. p(2) = 8 − 12 + 2k + 5 = 2k + 1 = 11 → k = 5.
D1. Using identity a³ + b³ = (a + b)(a² − ab + b²): 27x³ + 64y³ = (3x)³ + (4y)³ = (3x + 4y)(9x² − 12xy + 16y²).
D2. If (x + a) is a factor: p₁(−a) = a² − pa + q = 0 → q = pa − a². p₂(−a) = a² − ma + n = 0 → n = ma − a². Subtracting: n − q = (ma − a²) − (pa − a²) = a(m − p). Therefore a = (n − q)/(m − p).
D3. Try x = 1: 1 − 6 + 11 − 6 = 0 ✓. Divide by (x − 1) to get x² − 5x + 6 = (x − 2)(x − 3). So x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3). Verify: expand back.
D4. 103³ = (100 + 3)³ = 100³ + 3(100²)(3) + 3(100)(9) + 27 = 1000000 + 90000 + 2700 + 27 = 1,092,727.
D5. (x + 1/x)² = x² + 2 + 1/x² → 36 = x² + 1/x² + 2 → x² + 1/x² = 34.
E1. No. The polynomial 5 (a constant) has degree 0, not 5. The degree is the highest power of the variable, and since there's no variable, the degree is 0. — Why students get this wrong: They confuse the value of the constant with the degree of the polynomial.
E2. After simplification, (x² − 1)/(x + 1) = (x − 1)(x + 1)/(x + 1) = x − 1, which is a polynomial. However, the original expression is not defined at x = −1, so strictly it is a rational expression, not a polynomial. — Why students get this wrong: They simplify and assume the result is always a polynomial, forgetting the domain restriction at x = −1.